50. Pow(x, n)#34
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fuga-98
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Jun 13, 2025
| * しばらく悩んでそもそもアルゴリズムがよくわかってなかったので解答を見て `Solution2AC` を書いた。 | ||
| * n < 0 がありうるのを見逃していした | ||
| * n % 2 の判定を 1 回しかしていなかったり、result *= result (base *= base ではなく) していたり、そもそもアルゴリズムをちゃんと理解していなかった | ||
| * LeetCode の書き方は x, n を破壊しているんだけど、変数名も気に食わないし、base と exponent にした。 |
fuga-98
reviewed
Jun 13, 2025
| base = x | ||
| exponent = n | ||
| result = 1 | ||
| while exponent != 0: |
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exponent > 0のほうが好みです。
なんか、ちゃんと終了してくれそうなので。
Fuminiton
reviewed
Jun 14, 2025
| class Solution1WA: | ||
| def myPow(self, x: float, n: int) -> float: | ||
| if n == 0: | ||
| return 1 |
| class Solution: | ||
| def myPow(self, x: float, n: int) -> float: | ||
| if n < 0: | ||
| x = 1.0 / x |
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n<0で分母に0がくる可能性があるので、x=0の時を早期リターンしても良いかと思いました。
oda
reviewed
Jun 16, 2025
| ### step3 | ||
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| * 2:30 | ||
| * 書いていて思ったが step2 のビットで腹落ちしていないのって、変数名の定義の他には、多分こちらの (step1 ベースの) 解答でも当てはまっていて、result を、exponent % 2 == 1 のときにしかアップデートしていないからな気がする？ |
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これは何も分かっていない感じがします。myMultiply にして *= を += に置き換えたら分かりますか?
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50. Pow(x, n)
https://leetcode.com/problems/powx-n/